2
$\begingroup$

Consider the eigenvalue equation for the $\hat{l}_z$ angular momentum operator: $$\hat{l}_zY_{lm_l}(\theta,\phi)=m\hbar Y_{lm_l}(\theta,\phi)$$ with separable solution $$Y_{lm_l}(\theta,\phi)=\Theta_{lm_l}(\phi)\exp(im\phi)$$

I am unsure about the following passage from Gasiorowicz's Quantum Physics:

It is sometimes argued that since […] a transformation $\phi\to\phi+2\pi$ leaves the system invariant, it is necessary that $\exp(2\pi im)=1$, so that $m$ is an integer.

This is not quite correct, since the quantities that enter into physical observables are of the type $\int_0^{2\pi}d\phi\Psi_1^*(\phi)A\Psi_2(\phi)$, with wavefunctions $\Psi(\phi)$ of the form $$\Psi(\phi)=\sum_{m=-\infty}^\infty C_m\exp(im\phi).$$

If we require that these arbitrary wavepackets do not change under the transformation $\phi\to\phi+2\pi$, then we are led to the conclusion that the most general allowed values of $m$ are $m=c+$ integer, where $c$ is a constant.

I recognise the integral in the passage as the expectation value of an observable $A$. I understand why the boundary conditions imply $m$ must be an integer; what I don't understand is how/why is $m$ allowed, in principle, to be equal to an integer plus a constant. While it may seem an irrelevant issue, I suspect that when $c=1/2$ then $m$ is half-integer and therefore an eigenvalue of spin, so I would like to understand exactly where this constant factor comes in. If I just plug in $m+$ constant in the equation and then I transform from $\phi$ to $\phi+2\pi$ I seem to break the boundary condition for single-valuedness.

$\endgroup$
1
$\begingroup$

Wavefunctions are not observable and and only unique up to a complex phase (i.e. $e^{ic \phi}$), since any complex phase will be equal to one if multiplied with its conjugate. In other words, you can multiply your wavefunction with any $e^{ic \phi}$, and it's still considered the same wavefunction! That said, $c$ can be any numbers, not necessary $1/2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.